Answer to Question #242997 in Algebra for ush

Question

A system of equations is given below, 𝑡𝑥 + 2𝑦 + 3𝑧 = 𝑎 2𝑥 + 3𝑦 − 𝑡𝑧 = 𝑏 3𝑥 + 5𝑦 + (𝑡 + 1)𝑧 = 𝑐 Where 𝑡 is an integer and 𝑎, 𝑏, 𝑐 are real constants. The system does not have a unique solution, but it is consistent. Show that 𝑎 + 𝑏 = 𝑐.

Answer 1

System of equations are: 𝑡𝑥 + 2𝑦 + 3𝑧 = 𝑎, 2𝑥 + 3𝑦 − 𝑡𝑧 = 𝑏, 3𝑥 + 5𝑦 + (𝑡 + 1)𝑧 = 𝑐.

Now, $\begin{vmatrix} t & 2 & 3 \\ 2 & 3 &-t \\ 3 & 5 & (t+1) \end{vmatrix}=0$ [ System does not have a unique solution, but it is consistent ]

$t[3t+3+5t]-2[2t+2+3t]+3[10-9]=0\\ \Rightarrow 8t^2+3t-10t-4+3=0\\ \Rightarrow 8t^2-7t-1=0\\ \Rightarrow 8t^2-8t+t-1=0\\ \Rightarrow (8t+1)(t-1)=0\\ \therefore t=1,\frac{-1}{8}$

But ""t"" is an integer.

So, $t=1$

Now, system of equations are: 𝑥 + 2𝑦 + 3𝑧 = 𝑎, 2𝑥 + 3𝑦 − 𝑧 = 𝑏, 3𝑥 + 5𝑦 + 2𝑧 = 𝑐.

So, adding first 2 equations, we get: 3𝑥 + 5𝑦 + 2𝑧 = 𝑎+𝑏

Comparing this equation with third equation, we get:

$a+b=c$

Hence Proved.